How to find the last $4$ digits of $2^{2020}$ with Euler

Question

I tried to find the $4$ digits of $2^{2020}$ with Euler.

i find without Euler the last digits (6) , but i dont know how to find it using Euler . its look very simple but im getting trouble on it...

Answer 1

I would like to preface that this is an admittedly long-winded way of solving such a problem. There may indeed be savvier ways but I feel this is the most elementary.

The problem boils down to finding an $x$ such that $2^{2020}\equiv x\bmod 10000$.

To do something like this, I would use the Chinese remainder theorem.

Note that $10000=10^4=(2\times 5)^4=2^4\times 5^4$, so we can decompose this congruence like so $$2^{2020}\equiv a\bmod2^4,\quad2^{2020}\equiv b\bmod 5^4,$$ and then combine accordingly.

So let's consider the $2^4$ congruence first. Well, this is trivial - $2^4\equiv 0\bmod 2^4$, so $2^{2020}\equiv (2^4)^{505}\equiv 0\bmod 2^4$.

Now, let's look at the $5^4$ congruence. For this, we use Euler's theorem, noting that $5^4$ and $2$ are coprime, so $2^{\varphi(5^4)}\equiv 1\bmod 5^4$. We can work out $\varphi(625)$ like so $$\varphi(625)=625\Big(1-\frac{1}{5}\Big)=500.$$ Then $$2^{2020}\equiv(2^{500})^4\times2^{20}\equiv 2^{20}\equiv 1048576\equiv 451\bmod 5^4.$$ So, we have $2^{2020}\equiv 0\bmod 2^4$ and $2^{2020}\equiv 451\bmod 5^4$. Now, the second congruence implies there exists an integer $k$ such that $2^{2020}=625k+451$. Then, this implies that $625k+451\equiv 0\bmod2^4\Rightarrow 625k\equiv 13\bmod 2^4\Rightarrow k\equiv 13\bmod 2^4$. This means that there exists an integer $r$ such that $k=16r+13$. Substituting this back into the other equation involving $k$, we conclude that $$k=625(16r+13)+451=10000r+8125+451=10000r+8576.$$ Thus, the last four digits are $8576$.